$x^{x^2} + x^{x^8} =?$ Given $x^{x^4} = 4$

Question

If x is any complex number such that $x^{x^4} = 4$ , then find all the possible values of :

$x^{x^2} + x^{x^8}$

First, I used laws of exponents to give $18$ as answer. However , I realised that I've misused it. Further I used logarithms which yielded

$$ x^{x^2} + x^{x^8} = 4^{1/x^2} + 4^{x^2} $$

After this, I am stuck and can't proceed further. By hit and trial, $x = \sqrt{2}$ seems to be one of the solution. Any help is appreciated.

Answer 1

This answer only works (based on the original question) on the set of real numbers.

Hint:

\begin{align} x^{x^4}=4&\implies \left(x^4\right)^{x^4}=4^4\\ &\implies x^4=4\\ &\implies x=\pm\sqrt 2=\pm 2^{\frac 12}.\end{align}

---

Justification about the step $$\left(x^4\right)^{x^4}=4^4\implies x^4=4$$

We know that if $x\ge0$ and $x,y\in\mathbb R$ then the equation $ye^y=x$ has exactly one solution $W_0(x)$. Hence, we have

$$x^4\ln x^4=\ln x^4 e^{\ln x^4}=4\ln 4\ge 0$$

So, the fact $x^4=4$ is an only possible solution.

Answer 2

The value of $x^{x^2}+x^{x^8}$ is not uniquely determined by the condition $x^{x^4}=4$ if we work in the complex numbers, as can be observed by considering $x=\sqrt 2$ and $x=i\sqrt 2$. So I'd say that the problem is ill-posed. The original problem is probably intended to be posed in the context of real numbers.

Answer 3

update: I've seen now that we can find infinitely many solutions, adressable by a key $(p,q)$ where $p,q \in \mathbb Z$. Some keys produces duplicates of the four values $\sqrt2 \cdot \{1,î,-1,-î\}$. I do not yet know whether there are more duplicates, and also not yet whether there are further solutions besides that defined set. Details see below.

---

(Original post)
I've got the following $12$ solutions for $x \in \mathbb C$:

   k                                   x                                      x^x^4
------------------------------------------------------------------ -----------------------
   0  1.41421356237309504880168872421                                       4.000000000
   1  1.73694151125389186600122272542 + 0.425762209627875509859692133261*I  4.000000000
   2  1.94780723984600666553862444404 + 0.545039374401979794335694544212*I  4.000000000
   3  2.09437773294377459599457685403 + 0.616893780731956963391891512255*I  4.000000000
   4  2.20894368669214706385956705826 + 0.669622647771460774484724980808*I  4.000000000
   5  2.30405416343400277944984245993 + 0.711838306901809430540522438479*I  4.000000000
   6  2.38595034118298196996387646734 + 0.747336439812107913017309003950*I  4.000000000
   7  2.45821823819247099509976444780 + 0.778139236849140680600612225044*I  4.000000000
   8  2.52312195993007435475083523369 + 0.805457990541472410305864513590*I  4.000000000
   9  2.58218933276750739100666975088 + 0.830078738731896713946415742162*I  4.000000000
  10  2.63650370500133271312584396238 + 0.852542181022368142956408377652*I  4.000000000
  11  2.68686330065792127959502336814 + 0.873237139433454959225648272548*I  4.000000000

_{Note 1: Here $x$ are computed to 100 dec digits internally; only 30 digits are shown
Note 2: This results are found using branches of the complex logarithm. Infinitely many similar solutions can be given.}

Solutions to your question in title are as follows (the same $x_k$ are used):

  k             x                   x^x^4          x^x^2 + x^x^8 
   0  1.414213562                 4.000000000    258.0000000
   1  1.736941511+0.4257622096*I  4.000000000    0.1070042320+3.641589540*I
   2  1.947807240+0.5450393744*I  4.000000000   -5.066165533+4.197230968*I
   3  2.094377733+0.6168937807*I  4.000000000   -10.88664880-0.2571870638*I
   4  2.208943687+0.6696226478*I  4.000000000   -13.69826153-10.13887251*I
   5  2.304054163+0.7118383069*I  4.000000000   -9.653328410-23.74966607*I
   6  2.385950341+0.7473364398*I  4.000000000    4.262811997-37.18679992*I
   7  2.458218238+0.7781392368*I  4.000000000    29.16692177-44.70029785*I
   8  2.523121960+0.8054579905*I  4.000000000    63.34861041-39.50053796*I
   9  2.582189333+0.8300787387*I  4.000000000    101.6469692-14.96732788*I
  10  2.636503705+0.8525421810*I  4.000000000    135.2862732+33.84474321*I
  11  2.686863301+0.8732371394*I  4.000000000    152.2904794+108.5620067*I

The orbit in the complex plane for the first 32 solutions looks like this:

For the first 12 solutions only, to see more detail:

If the spiral of the orbit is scaled using the log of the distances of the values to the origin, then the resulting shape suggests a smooth estimate for the distances:

---

update: By revision of my procedure I can now generate infinitely infinite sets of solutions $x^{x^4} = 4, \qquad x \in \mathbb C$; here I show the sets "$\text{roots}(-15..16,-12..12)$" where in the picture the subsets $\text{roots}(-15..16,q)$ (using $q=-12..12$) get the same color for same $\pm q$: It may be interesting,that the values $x \in \sqrt2\cdot\{1,î,-1,-î\}$ occur repeatedly in the sets defined by $q$, even if at different indexes $p$.